Question

$C$ is a closed path in the $z-$plane given by $|z|=3$. The value of the integral ${\oint }_C\left(\frac{z^2-z+4j}{z+2j}\right)dz$ is

• A. $-4\pi (1+j2)$
• B. $4\pi (3+j2)$
• C. $-4\pi (3+j2)$
• D. $4\pi (1-j2)$

Answer 1

The correct option is C $-4\pi (3+j2)$

Method I:

$f\left(z\right)=\left(\frac{z^2-z+4j}{z+2j}\right)$ so pole is $z=-2j$
$c:|z|=3$



So pole lies inside ${}^{\prime }{c}^{\prime }$
$R=\underset{(z=-2j)}{Resf\left(z\right)}=\underset{z\to 2j}{lim}\left[\right(z+2j\left)f\right(z\left)\right]$
$={lim}_{z\to -2j}(z^2-z+4j)=-4+6j$
By Residue theorem,
${\oint }_{c}f\left(z\right)dz=2\pi j$ (Sum of residues)
$=2\pi j(-4+6j)=-12\pi -8\pi j$
$=-4\pi (3+2j)$

Method II:
$I={\oint }_C\frac{z^2-z+4j}{z+2j}dz,|Z|=3$
$z=-2j$ is singular point which lies inside $|z|=3$.
By cauchy integral formula
${\oint }_C\frac{z^2-z+4j}{z+2j}dz=2\pi j(-2j)$
$=2\pi j(4j^2+2j+4j)$
$=2\pi j(6j-4)$
$=2\pi (+6j^2-4j$)
$=-4\pi (3+2j)$